Find all field homomorphisms $f:\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$.
We know that: $ \mathbb{Q}(\sqrt{2}) = \text{span}_\mathbb{Q} \{ 1, \sqrt{2} \} = \{ a+\sqrt{2}b \,|\, a,b\in \mathbb{Q} \} $
And that a field homomorphism fulfills the following criteria:
- $\forall x \in \mathbb{Q}(\sqrt{2}): \forall \lambda \in \mathbb{Q}: f(\lambda x)=\lambda f(x)$
- $\forall x, x' \in \mathbb{Q}(\sqrt{2}): f(x+x')=f(x)+f(x')$
- $\forall x, x' \in \mathbb{Q}(\sqrt{2}): f(xx')=f(x)f(x')$
Let $a,b,a',b' \in \mathbb{Q}$, then it follows that $x=a+\sqrt{2}b$ and $x'=a'+\sqrt{2}b'$ are in $\mathbb{Q}(\sqrt{2})$.
Which means that criteria 3 should apply: \begin{equation} f\big((a+\sqrt{2}b)(a'+\sqrt{2}b')\big)=f\big(a+\sqrt{2}b\big)f\big(a'+\sqrt{2}b'\big) \end{equation}
Now, multiply everything out and then use the $\mathbb{Q}$-linearity of $f$ to get to the following equation:
\begin{equation} 0=aa'f(1)(1-f(1))+ab'f(\sqrt{2})(1-f(1))+a'bf(\sqrt{2})(1-f(1))+bb'(2f(1)-f(\sqrt{2})^2) \end{equation}
Now, set $a,a',b,b'$ to either $0$ or $1$ so that only one of the terms remains, which leads to these equations: \begin{align} 0 &= f(1)(1-f(1)) \\ 0 &= f(\sqrt{2})(1-f(1)) \\ 0 &= f(\sqrt{2})(1-f(1)) \\ 0 &= 2f(1)-f(\sqrt{2})^2 \end{align}
So the final answer should be: \begin{align} \{ f: \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}) \, | \, &(f(1)=f(\sqrt{2})=0) \vee \\ & (f(1)=f(\sqrt{2})=1) \vee \\ & (f(1)=-1 \wedge f(\sqrt{2})=1) \} \end{align}
Is my reasoning correct?